Fun with maths: Let x = ln(sec(y) + tan(y). Show that sec(y) = cosh (x).?
Remember that cosh(x) = (1/2) * (e^(x) + e^(-x)) ..1 and as sec^2 y- tan ^2 y = 1 so 1/(sec y + tan y) = sec y - tan y ..2 x = ln(sec(y) + tan(y)) so e^x = sec(y) + tan(y) ...3 so e^-x = 1/( sec y + tan y) = sec y - tan y .. 4 ( from 2) add (3) and (4) to...